Page 132 - Solutions Manual to accompany Electric Machinery Fundamentals

E 587 V
I  A 2 I   4 A  4.89 A
F 2 F 1
E A 1 480 V

(b) When the load is removed from the motor the magnitude of E remains unchanged but the torque
A
0
angle goes to  . The resulting armature current is
V  E 480 0 V 587 0





I   A   178.3 90 A
A
jX S j 0.6 
5-3. A 230-V, 50 Hz, two-pole synchronous motor draws 40 A from the line at unity power factor and full
load. Assuming that the motor is lossless, answer the following questions:
(a) What is the output torque of this motor? Express the answer both in newton-meters and in pound-
feet.
(b) What must be done to change the power factor to 0.85 leading? Explain your answer, using
phasor diagrams.
(c) What will the magnitude of the line current be if the power factor is adjusted to 0.85 leading?

SOLUTION

(a) If this motor is assumed lossless, then the input power is equal to the output power. The input
power to this motor is

P  IN 3V I cos    3 230 V 40 A 1.0
  15.93 kW
TL
The rotational speed of the motor is
120 f  120 50 Hz
n  se   1500 r/min
m
P 4
The output torque would be
P 15.93 kW

  OUT   101.4 N m
LOAD
 m  1500 r/min  1 min   2 rad  
  60 s     1 r  
In English units,
7.04 P  7.04 15.93 kW 
 LOAD  OUT   74.8 lb ft 
m n 1500 r/min
(b) To change the motor’s power factor to 0.8 leading, its field current must be increased. Since the
power supplied to the load is independent of the field current level, an increase in field current increases
E while keeping the distance E A sin  constant. This increase in E A changes the angle of the
A
current I A , eventually causing it to reach a power factor of 0.8 leading.

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